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Am 20.07.2024 um 15:30 schrieb Richard Damon:
> On 7/20/24 8:47 AM, WM wrote:

>> My theorem: X is left-hand side of every x > 0 <==> X is left-hand 
>> side of (0, oo).

Of course he has forgotten to state the "domain" of his "theorem" 
explicitely. But it's reasonable zu assume that "X" and "x" here are 
used as as varibles for real numbers.

>> Find a counter example or accept it.

> 
> But no [...] x can BE that left hand side of (0, oo) because if such 
> an x did exist then x/2 would be outside the interval, but also positive.

0 can't be "left hand side of (0, oo)" (WM)?

Hint: WM uses non-standard terminology here. "X is left hand side of (0, 
oo)" means: Ax e (0, oo): X < x. We might abbreviate this with X < (0, oo).

If we state his "X is left-hand side of every x > 0" in the usual 
formal/symbolic language (which is uses in the context of set theory) we 
get: "Ax e IR, x > 0: X < x".

Now clearly: Ax(x e IR & x > 0 <-> x e (0, oo)). Hence his "theorem" is 
just a simple tautology (based on a simple definition):

Def.: X < (0, oo) :<-> Ax e (0, oo): X < x

Now:

       Ax e IR, x > 0: X < x <-> Ax e (0, oo): X < x

since by definition (of an intervall) Ax(x e IR & x > 0 <-> x e (0, oo).

Hence (with def. from above):

       Ax e IR, x > 0: X < x <-> X < (0, oo) .

A trivial "result".

Though I'd prefere it to state it the following way:

       AX e IR: (Ax e IR, x > 0: X < x <-> X < (0, oo)) .

_________________________________________

Now we may, say, consider the set {X e IR : X < (0, oo)}. Then we get:

       {X e IR : X < (0, oo)} = (-oo, 0].

Using an additional "extension" of "<" we might even state this result as

       (-oo, 0] < (0, oo)

Meaning: Ax,y e IR: x e (-oo, 0] & y e (0, oo) -> x < y.